Question

a walk (or row) is miles (per hour) in still water. the river (or the wind) (or the current) is miles per hour. when upstream (or against the current/wind) and downstream (or with the current/wind) in the same amount of time, she (or he) can row (or walk) 25 miles downstream (or with) and 15 miles upstream (or against) by solving \\(\frac{25}{x + 4} = \frac{15}{x - 4}\\).\
\bigcirc no solution\
\bigcirc 9 mph\
\bigcirc 3 mph\
\bigcirc 12 mph\
\bigcirc 15 mph

Explanation:

Step1: Cross - multiply the proportion

Given the equation \(\frac{25}{x + 3}=\frac{15}{x-3}\), cross - multiplying (which is based on the property of proportions: if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d=b\times c\)) gives us \(25(x - 3)=15(x + 3)\).

Step2: Expand both sides

Using the distributive property \(a(b + c)=ab+ac\), we expand the left - hand side: \(25x-75\) and the right - hand side: \(15x + 45\). So the equation becomes \(25x-75 = 15x+45\).

Step3: Isolate the variable \(x\)

Subtract \(15x\) from both sides: \(25x-15x-75=15x - 15x+45\), which simplifies to \(10x-75 = 45\). Then add 75 to both sides: \(10x-75 + 75=45 + 75\), so \(10x=120\).

Step4: Solve for \(x\)

Divide both sides by 10: \(\frac{10x}{10}=\frac{120}{10}\), so \(x = 12\). But wait, the question is about the speed of the river current? Wait, maybe there was a mis - reading. Wait, the original problem is about a boat: let the speed of the boat in still water be \(x\) and the speed of the current be \(c\). When going downstream, the speed is \(x + c\), and upstream is \(x - c\). The time taken to go 25 miles downstream and 15 miles upstream is the same. But from the equation \(\frac{25}{x + 3}=\frac{15}{x-3}\), maybe \(x\) is the speed of the boat and 3 is the speed of the current? Wait, no, we solved for \(x\) and got \(x = 12\). Wait, maybe the speed of the current is 3? Wait, no, let's re - check. Wait, if we assume that the speed of the boat in still water is \(x\) and the current is \(c\), and the time \(t=\frac{d}{v}\) (distance divided by speed). If the time for 25 miles downstream (speed \(x + c\)) and 15 miles upstream (speed \(x - c\)) is the same. If we got \(x = 12\), and if we substitute back into \(x + 3\) and \(x - 3\), downstream speed is \(15\) mph and upstream speed is \(9\) mph. But the question is about the speed of the current? Wait, maybe the equation was set up with the current speed as 3? No, wait, our solution for \(x\) (the speed of the boat) is 12, but the options are 3, 9, 12, 15. Wait, maybe the question is asking for the speed of the current? Wait, no, let's re - examine the equation. If we made a mistake in the problem interpretation. Wait, if the equation is \(\frac{25}{v_{down}}=\frac{15}{v_{up}}\), and \(v_{down}=x + 3\), \(v_{up}=x - 3\), and we found \(x = 12\), then the speed of the current is 3? Wait, no, the options include 3 mph. Wait, maybe the question is asking for the speed of the current. Let's check: if \(x = 12\) (speed of boat in still water), then downstream speed is \(12 + 3=15\) mph, upstream speed is \(12-3 = 9\) mph. Time for 25 miles downstream: \(\frac{25}{15}=\frac{5}{3}\) hours, time for 15 miles upstream: \(\frac{15}{9}=\frac{5}{3}\) hours. So the speed of the current is 3 mph? Wait, no, the options have 3 mph as an option. Wait, maybe the question was to find the speed of the current, and the equation was set up with the current speed as \(c\), and the boat's speed as \(x\), and we solved for \(c\)? Wait, no, our solution for \(x\) (boat speed) is 12, but the option 3 is there. Wait, maybe I misread the equation. Let's re - do the equation:

\(\frac{25}{x + 3}=\frac{15}{x-3}\)

Cross - multiply: \(25(x - 3)=15(x + 3)\)

\(25x-75 = 15x + 45\)

\(25x-15x=45 + 75\)

\(10x=120\)

\(x = 12\). But if the question is about the speed of the current, and the equation was set up with the current speed as 3, then maybe the answer is 3 mph? Wait, the options are: No solution, 9 mph, 15 mph, 3 mph, 12 mph. Wait, when we solve the equation, we get \(x = 12\), but maybe \(x\) is the speed of the current? No, that doesn't make sense. Wait, maybe the original problem is: A boat can row \(x\) miles per hour in still water. It takes the same amount of time to row 25 miles downstream (where the speed is \(x + 3\)) as it does to row 15 miles upstream (where the speed is \(x - 3\)). Find the speed of the current (which is 3)? No, the speed of the current is 3? Wait, no, the value of \(x\) (boat speed) is 12, and the current speed is 3. But the options include 3 mph. Wait, maybe the question was to find the speed of the current, and the equation was set up with the current speed as \(c = 3\). Wait, I think I made a mistake in the problem interpretation. Let's check the options. The options are: No solution, 9 mph, 15 mph, 3 mph, 12 mph. When we solve the equation \(\frac{25}{x + 3}=\frac{15}{x-3}\), we get \(x = 12\). But if the question is about the speed of the current, and the equation was set up with the current speed as 3, then the answer is 3 mph? Wait, no, the boat's speed is 12, current is 3. But the equation was solved for \(x\) (boat speed) as 12. But the options have 3 as an option. Wait, maybe the equation was \(\frac{25}{v_{down}}=\frac{15}{v_{up}}\), and \(v_{down}=15\), \(v_{up}=9\), and the current speed is \(\frac{15 - 9}{2}=3\). Yes, because downstream speed = boat speed + current speed, upstream speed = boat speed - current speed. So if downstream speed is 15 and upstream is 9, then boat speed \(b=\frac{15 + 9}{2}=12\), current speed \(c=\frac{15 - 9}{2}=3\). So the speed of the current is 3 mph.

Answer:

3 mph